Here you will learn what is the differentiation of cosecx and its proof by using first principle.<\/p>\n
Let’s begin –<\/p>\n
\nThe differentiation of cosecx with respect to x is -cosecx.cotx<\/p>\n
i.e. \\(d\\over dx\\) (cosecx) = -cosecx.cotx<\/p>\n<\/blockquote>\n
Proof Using First Principle :<\/h2>\n
\nLet f(x) = cosec x. Then, f(x + h) = cosec(x + h)<\/p>\n
\\(\\therefore\\)\u00a0 \u00a0\\(d\\over dx\\)(f(x)) = \\(lim_{h\\to 0}\\) \\(f(x + h) – f(x)\\over h\\)<\/p>\n
\\(d\\over dx\\)(f(x)) = \\(lim_{h\\to 0}\\) \\(cosec(x + h) – cosec x\\over h\\)<\/p>\n
\\(\\implies\\)\u00a0 \\(d\\over dx\\)(f(x)) = \\(lim_{h\\to 0}\\) \\({1\\over sin(x + h)} – {1\\over sin x}\\over h\\)<\/p>\n
\\(\\implies\\)\u00a0 \\(d\\over dx\\)(f(x)) = \\(lim_{h\\to 0}\\) \\(sin x – sin(x + h)\\over h sin x sin(x +h)\\)<\/p>\n
By using trigonometry formula,<\/p>\n
[sin C – sin D = \\(2sin ({C – D\\over 2})cos ({C + D\\over 2})\\)]<\/p>\n
\\(\\implies\\) \\(d\\over dx\\)(f(x)) = \\(lim_{h\\to 0}\\) \\(2sin ({x – x – h\\over 2})cos({x + x + h\\over 2})\\over h sin x sin (x + h)\\)<\/p>\n
\\(\\implies\\) \\(d\\over dx\\)(f(x)) = \\(lim_{h\\to 0}\\) \\(2sin ({-h\\over 2})cos({x + h\/2})\\over h sin x sin (x + h)\\)<\/p>\n
\\(\\implies\\) \\(d\\over dx\\)(f(x)) = -\\(lim_{h\\to 0}\\) \\(cos ({x + h\/2})\\over sin x sin(x + h)\\).\\(lim_{h\\to 0}\\) \\(sin(h\/2)\\over (h\/2)\\)<\/p>\n
because, [\\(lim_{h\\to 0}\\)\\(sin(h\/2)\\over (h\/2)\\) = 1]<\/p>\n
\\(\\implies\\) \\(d\\over dx\\)(f(x)) = -\\(cos x\\over sin x sin x\\)(1) = -cot x cosec x<\/p>\n
Hence, \\(d\\over dx\\) (cosec x) = =cosecx.cotx<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : What is the differentiation of cosec x + x with respect to x?<\/p>\n
Solution<\/strong> <\/span>: Let y = cosec x + x<\/p>\n
\\(d\\over dx\\)(y) = \\(d\\over dx\\)(cosec x + x)<\/p>\n
\\(\\implies\\) \\(d\\over dx\\)(y) = \\(d\\over dx\\)(cosec x) + \\(d\\over dx\\)(x)<\/p>\n
By using cosecx differentiation we get,<\/p>\n
\\(\\implies\\) \\(d\\over dx\\)(y) = -cosec x cot x + 1<\/p>\n
Hence, \\(d\\over dx\\)(sec x + x) = -cosec x cot x + 1<\/p>\n
Example<\/strong> <\/span>: What is the differentiation of \\(cosec\\sqrt{x}\\) with respect to x?<\/p>\n
Solution<\/strong><\/span> : Let y = \\(cosec\\sqrt{x}\\)<\/p>\n
\\(d\\over dx\\)(y) = \\(d\\over dx\\)(\\(cosec\\sqrt{x}\\))<\/p>\n
By using chain rule we get,<\/p>\n
\\(\\implies\\) \\(d\\over dx\\)(y) = \\(1\\over 2\\sqrt{x}\\)(\\(-cosec \\sqrt{x}.cot\\sqrt{x}\\))<\/p>\n
Hence, \\(d\\over dx\\)(\\(cosec\\sqrt{x}\\)) = -\\(1\\over 2\\sqrt{x}\\)(\\(cosec \\sqrt{x}.cot\\sqrt{x}\\))<\/p>\n
\nRelated Questions<\/h3>\n
What is the Differentiation of cosec inverse x ?<\/a><\/p>\n